Tìm nghiệm: `{(3x-1)(x+3)}/3``-``{2x^2+1}/2``=``15/2`

`{( 3x - 1 ) ( x + 3 )}/3 - {2x^2 + 1}/2 = 15/2`

`⇔ {3x^2 + 8x - 3}/3 - {2x^2 + 1}/2 = 15/2`

`⇔ {6x^2 + 16x - 6 - 6x^2 - 3 }/6 = 15/2`

`⇔ {16x - 6 - 3 }/6 = 15/2`

`⇔ 16x - 9  = 15/2`

`⇔ 16x - 9 = 45`

`⇔ 16x = 45 + 9`

`⇔ 16x = 54`

`⇔ x = 27/8`

Vậy `S ∈ { 27/8}`

Đáp án:

 $\dfrac{(3x-1).(x+3)}{3}-\dfrac{2x^2+1}{2}=\dfrac{15}{2}$

$⇔ \dfrac{2.(3x^2+8x-3)}{6} -\dfrac{3.(2x^2+1)}{6}= \dfrac{45}{6}$

$⇒ 6x^2+16x-6-6x²-3=45$

$⇔ 16x-9-45=0$

$⇔ 16x-54=0$

$⇔ 16x=54$

$⇔ x= \dfrac{27}{8}$

$\text{Vậy S=}$ $\left\{{\dfrac{{27}}{{8}}}\right\}$